Notice that in the center of mass frame: \(\mathbf {Z}=\mathbf {P}=0\) .
One could have instead built this action from a bottom-up approach based on symmetry argument, obtaining [1]}:
\(S_{source}&=&-E\int d\tau -\frac{1}{2}\int dx^\mu L^i\epsilon _{ijk}w_\mu ^{jk}-\int d\tau \sum _{l=2}^{\infty }\frac{1}{l!}I^L\nabla _{L-2}\mathcal {E}_{k_{l-1}k_l}\nonumber \\& &-\int d\tau \sum _{l=2}^{\infty }\frac{2l}{(l+1)!}J^L\nabla _{L-2}B_{k_{l-1}k_l} \ ,\) 
